How can you divide an angle into 3 equal angles? You may only use a straightedge and a compass to achieve this.

(This means : You have an angle A, you divide Angle A into 3 Angles B,C,D. And B=C=D=A/3)

Note: vohonam clarified that the problem actually only gives you a straightedge, not a ruler.

(In reply to eureka!!!!! part 2! by James)

I did not completely follow every step, but even assuming you got to step 6.10 (three equal lengths) correctly, you still failed to trisect the angle.

All you did was turn the original angle A into an isosceles triangle BAC and trisected side BC. Trisecting the base of an isoscelese triangle is not the same as trisecting the opposite angle.

Consider an isoscelese tiangle ABC where BC is the base. Trisect the base BC at points D and E, and draw AD and AE.

By the law of sines: [sin(BAD)]/(BD) = [sin(ABD)]/(AD) and [sin(EAD)]/(ED) = [sin(AED)]/(AD)

(BD) = (ED) because you constructed D and E that way. Now if we assume that angle BAD = angle EAD, then their sines would also be equal.

That would mean that [sin(ABD)]/(AD) = [sin(AED)]/(AD), and angles ABD and AED are equal. If two triangles have two equal angles, their third angles are also equal, so BDA = ADE and is therefore 90° so AD would be perpendicular to BC.

Similarly, AE would be perpendicular to BC. But there can only be one line perpendicular to BC which passes through A. So Angle DAE does not equal Angle BAD or angle EAC, and you did not trisect angle BAC.

Posted by TomM on 2002-10-26 03:57:23